My target is to solve next task: We have an triangle with points $A,B,C$. $AC$ => $\vec{\text{a}}$, $CB$ => $\vec{\text{b}}$ and $AB$ is divided by 6 equals parts with points $O, P, Q, R, S, T$. We must find width of $\vec{\text{CO}}$, $\vec{\text{CP}}$, $\vec{\text{CQ}}$, $\vec{\text{CR}}$, $\vec{\text{CS}}$ and $\vec{\text{CT}}$.
I know how to find vector $\vec{\text{CQ}}$, which is the middle of $AB$, by formula $\vec{\text{CQ}} = \frac{1}{2}(\vec{\text{a}} + \vec{\text{b}})$, but I do not know how to find others. Please explain. Thank you.
If segment $AB$ has to be divided in six equal parts, point $T$ has to be $B$ ($T=B)$ You can use the formula:
$$\vec a+(n/6)(\vec b-\vec a)$$
It works for every point in $AB$ E.g., with n=3 we get Q:
$$\vec{\text{CQ}}=\vec a+(3/6)(\vec b-\vec a)=(1/2)(\vec b+\vec a)$$
With $n=5$, $S$:
$$\vec{\text{CS}}=\vec a+(5/6)(\vec b-\vec a)$$